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Nepsesh

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About Nepsesh

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  1. Thanks so much for your answer, my error was in the (1-x^2) shell's 'height' assembly, ie missing the 1- from the limit. Much obliged, M.
  2. Please help. My attempt to rotate the volume about the x-axis of y=x^2, first by disk method, then by shell method produces two answers - where obviously it should be the same answer. The limits are from 1 to 0. 1. V=Pi int(x^2)^2 dx = Pi int(y^4)dx = Pi/5 = 0.62631 ft cbd. (Disk method using f(x) where y=x^2). 2. V=2Pi int y.(sqrt y) dy = 4xPi/5 = 2.51327 ft cbd. (Shell method using g(y) where x=sq rt y). Couldn't be simpler? I've spent hours trying to see what's wrong.
  3. Yes, very helpful. Now you mention it there's a lot of info online about this subject. I was previously aware of the awsome Rupert's Drops phenomenom yet did not connect the two. Thanks for your reply.
  4. Thanks for your most interesting and prompt replies. Much appreciated. M.
  5. A simple question that has always had me wondering. A broken pane invariably presents shards with smooth, perpendicular edges. A steel edge of 90 deg, even when precision ground and then honed, will not cut an unwary finger so readily as the bluntest profile in glass. The shape seems almost irrelevant to its keenness. Is it "molecular" or what? Does anybody know please?
  6. Thanks for this lead re Pippard. Have updated my printout with the small correction mentioned above. Nice to have corresponded with you, Mike_B.
  7. Thanks very much for your replies Xittenn, most helpful. I thought ellipse in part at least the answer as any one point on the helix at any time is touching the minor axis of the oblique cutting plane ellipse of a cylinder at the helix angle. Sufficiently true for a small arc but getting progressively worse without 'recutting' the plane through the cylinder repeatedly towards infinity. Bye, Mike. PS, I think a small elephant might be more outre to pull the rolled steel coil than a 4-wheel drive!
  8. Thanks for your reply Xittenn (or Atom?). You have the facts exactly. Yes the technique stated will work, brute force & oxy heat will shape almost anything, but its not subtle! A better technique but more impracticable would be to roll a circular coil of 1 rev of diameter "hypotenuse", then stretch it like a spring the distance of 12 risers to get the correct spiral. Between a tree & a 4-wheel drive vehicle might do it. Health & safety would stipulate 'eye protection' of course! This stretch would decrease the diameter back to the helix base dimension of Pi x OD & should fit exactly. But neither is the elegant solution I seek for my work table. (I do not have a 4-wheel drive either).. Mike. Sorry Xittenn, the hypotenuse L = root [(Pi OD)sqd + (height)sqd] not what you put.
  9. Hi everyone, I'm a new boy. I am a retired fabricator/welder, used to most steel structure work & the mathematics thereof. Most things can be drawn - projections, geometry, onto a steel worktable & in 2D. I now need to know how the helix of a banister for a spiral staircase can be drawn, say in 60 degree parts. This means one revolution of the stair would need 6 similar arcs welded together for the banister. The length of the whole banister (per rev) is the hypotenuse of the rt angle triangle made by the horizontal of Pi x staircase diameter & the vertical rise of the (usually) 12 steps in one revolution. This hypotenuse can then simply be divided by 6 to give the length of arc of two risers, ie, 60 deg of turn. Dividing the whole hypotenuse by Pi gives the working diameter of the arc just described. This can now be laid out on the table & the steel bent to this shape. Repeat this until the whole banister length has been acheived. But this procedure becomes increasingly inaccurate as the arc length is increased - the circle principle collapses to something like a part of an ellipse, or worse, a section of sine. Lots more metal bashing required! Is this simpler than I make it sound? No calculus please - its not much used in the welding industry... Many thanks for any help or comments, Mike.
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